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23-Chem-B1 Transport Phenomena · May 2014

Question 3 of 6: B1 — Heat generation in a hollow cylinder

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, May 2014 — 04-CHEM-B1 Transport Phenomena; 3 hours, open book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer); candidates attempt one from each section plus a fourth from any section (four of six marked, 25 marks each). The worked solutions below cover all six problems. A summary of the conservation equations (continuity, Navier–Stokes, energy, species) is provided as Appendix A in the paper and is used throughout.

Reference texts: R. S. Brodkey & H. C. Hershey, Transport Phenomena — A Unified Approach (McGraw-Hill) — the paper’s own reference and source of the appended conservation-equation tables; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — differential balances and diffusion with reaction; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — conduction with generation, composite walls, combined convection–radiation.

Question 3: B1 — Heat generation in a hollow cylinder

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Volumetric generation set by $2000$ W per metre of length; inner surface ($R_1=0.02$ m) insulated; outer surface ($R_2=0.04$ m) convecting to $T_\infty=25^\circ\text{C}$.

SymbolValue
Heat per length $\dot q'$$2000\ \text{W}\,\text{m}^{-1}$
Conductivity $k$$1.2\ \text{W}\,\text{m}^{-1}\text{K}^{-1}$
Convection $h$ / $T_\infty$$75\ \text{W}\,\text{m}^{-2}\text{K}^{-1}$ / $25^\circ\text{C}$
Radii $R_1$, $R_2$$0.02$ m, $0.04$ m

Find. (a) the radial temperature profile; (b) the maximum temperature and where it occurs.

insulated inner wall (T₁) generation q‴ R₁ = 0.02 m R₂ = 0.04 m h = 75 W/m²K T∞ = 25°C
Fig. B1: Hollow cylinder with uniform internal generation. The inner surface is insulated ($dT/dr=0$ at $R_1$), so all heat flows outward and leaves the outer surface by convection; the temperature therefore peaks at the inner wall.

Part (a) — temperature profile

The printed symbol $\dot q'''$ in the required expression is the volumetric generation rate (W m$^{-3}$) — the “W/m” on the figure label is the per-length load stated in the text — and below we write it simply as $\dot q$. We first tie it to the stated 2000 W per length, then integrate the cylindrical energy equation.

  1. Relate volumetric generation to the given line load. With uniform generation $\dot q$ over the annular cross-section, $\dot q'=\dot q\,\pi(R_2^2-R_1^2)$, so $$\dot q=\frac{\dot q'}{\pi(R_2^2-R_1^2)}=\frac{2000}{\pi(0.04^2-0.02^2)}=5.305\times10^{5}\ \text{W}\,\text{m}^{-3}.$$
  2. Reduce the energy equation. Steady, radial conduction with generation (Table A.3, cylindrical) gives $\dfrac{1}{r}\dfrac{d}{dr}\!\left(r\dfrac{dT}{dr}\right)+\dfrac{\dot q}{k}=0$, i.e. $\dfrac{d}{dr}\!\left(r\dfrac{dT}{dr}\right)=-\dfrac{\dot q}{k}r.$
  3. Integrate once and apply the insulated inner wall. $r\dfrac{dT}{dr}=-\dfrac{\dot q}{2k}r^2+C_1$. The insulated boundary $dT/dr=0$ at $r=R_1$ fixes $C_1=\dfrac{\dot q}{2k}R_1^2$, so $$\frac{dT}{dr}=\frac{\dot q}{2k}\left(\frac{R_1^2}{r}-r\right).$$
  4. Integrate again. $T(r)=\dfrac{\dot q}{2k}\left(R_1^2\ln r-\dfrac{r^2}{2}\right)+C_2.$
  5. Outer-surface energy balance sets the second constant. All generated heat crosses the outer surface: $\dot q'=h(2\pi R_2)\big(T(R_2)-T_\infty\big)$, giving $$T(R_2)=T_\infty+\frac{\dot q(R_2^2-R_1^2)}{2hR_2}.$$ Imposing this on the integrated profile and grouping the constants yields $$T(r)=\frac{\dot q}{2}\left[\frac{1}{k}\left(R_1^2\ln\frac{r}{R_2}+\frac{R_2^2-r^2}{2}\right)-\frac{1}{h}\left(\frac{R_1^2}{R_2}-R_2\right)\right]+T_\infty,$$ exactly the required expression. $\boxed{\,T(r)=\dfrac{\dot q}{2}\Big[\tfrac{1}{k}\big(R_1^2\ln\tfrac{r}{R_2}+\tfrac{R_2^2-r^2}{2}\big)-\tfrac{1}{h}\big(\tfrac{R_1^2}{R_2}-R_2\big)\Big]+T_\infty\,}$

Part (b) — maximum temperature

  1. Locate the maximum. From step 3, $dT/dr=\dfrac{\dot q}{2k}\big(R_1^2/r-r\big)$, which is zero at $r=R_1$ and negative for $r>R_1$. Temperature therefore decreases monotonically outward, so the maximum sits at the insulated inner wall, $r=R_1$.
  2. Evaluate the outer-surface temperature first. $T(R_2)=25+\dfrac{(5.305\times10^{5})(0.0012)}{2(75)(0.04)}=25+106.1=131.1^\circ\text{C}.$
  3. Evaluate at $r=R_1$. Using the profile, $$T_{max}=T(R_1)=\frac{\dot q}{2k}\left[R_1^2\ln\frac{R_1}{R_2}+\frac{R_2^2-R_1^2}{2}\right]+T(R_2).$$ With $\dot q/2k=2.210\times10^{5}$, $R_1^2\ln(0.5)=-2.773\times10^{-4}$ and $(R_2^2-R_1^2)/2=6.0\times10^{-4}$, the bracket is $3.227\times10^{-4}$, giving $71.3^\circ\text{C}$ above the outer surface: $$T_{max}=71.3+131.1=\boxed{202.4^\circ\text{C}}\quad(\text{at }r=R_1).$$
QuantityValue
Volumetric generation $\dot q$$5.305\times10^{5}\ \text{W}\,\text{m}^{-3}$
Outer-surface temperature $T(R_2)$$131.1^\circ\text{C}$
Maximum temperature (at $r=R_1$)$202.4^\circ\text{C}$